You shoot an arrow into the air. Two seconds later (2.00 {\rm s}) the arrow has gone straight upward to a height of 32.0 m above its launch point.

What is the initial velocity and how long did it take for the arrow to reach 16.0 m above the launch point?

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To find the initial velocity of the arrow, we can use the equation for vertical motion:

h = v0*t - (1/2)*g*t^2

where:
h = height (32.0 m)
v0 = initial velocity
t = time (2.00 s)
g = acceleration due to gravity (-9.8 m/s^2)

Substituting the given values into the equation, we get:

32.0 = v0*(2.00) - (1/2)*(9.8)*(2.00)^2

Simplifying the equation, we have:

32.0 = 2.00v0 - 19.6

Rearranging the equation, we find:

2.00v0 = 32.0 + 19.6
2.00v0 = 51.6

Dividing both sides by 2.00, we get the initial velocity:

v0 = 51.6 / 2.00
v0 = 25.8 m/s

The initial velocity of the arrow is 25.8 m/s.

To determine how long it took for the arrow to reach 16.0 m above the launch point, we can use the same equation. We will substitute the new height (16.0 m) and solve for the time (t):

16.0 = 25.8t - (1/2)*(9.8)*t^2

Simplifying the equation, we have:

16.0 = 25.8t - 4.9t^2

Rearranging the equation, we find:

4.9t^2 - 25.8t + 16.0 = 0

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 4.9, b = -25.8, and c = 16.0. Substituting these values into the quadratic formula, we get:

t = (-(-25.8) ± √((-25.8)^2 - 4*(4.9)*(16.0))) / (2*(4.9))

Simplifying further, we have:

t = (25.8 ± √(665.64 - 313.6)) / 9.8
t = (25.8 ± √(352.04)) / 9.8

t ≈ (25.8 ± 18.77) / 9.8

The two solutions for t are:

t ≈ (25.8 + 18.77) / 9.8 ≈ 4.53 s
t ≈ (25.8 - 18.77) / 9.8 ≈ 0.71 s

Therefore, it took approximately 4.53 seconds and 0.71 seconds for the arrow to reach 16.0 m above the launch point.